题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4578
四种操作:查询、加法、乘法、改数。应该是需要维护三个lazy标记,然后就是套路了。查询是区间内所有的数的p次幂然后再求和,这个p只有三个值(1,2,3),直接维护三棵线段树,分别是1 2 3次幂。
注意延迟标记的时候,如果有改数,那之前的加法和乘法就可以不用做了。在更新乘法的时候,如果有加法存在,那加法的标记应该更新,乘一下乘法的数,因为(a+b)*c = a*c+b*c,父亲是a+b,儿子是a和b。
pushdown的顺序就是:改数、乘法、加法。三种询问好处理,(a+b)^2和(a+b)^3展开就行了。
这是我做过的比较复杂的线段树了。
1 /* 2 ┓┏┓┏┓┃キリキリ♂ mind! 3 ┛┗┛┗┛┃\○/ 4 ┓┏┓┏┓┃ / 5 ┛┗┛┗┛┃ノ) 6 ┓┏┓┏┓┃ 7 ┛┗┛┗┛┃ 8 ┓┏┓┏┓┃ 9 ┛┗┛┗┛┃ 10 ┓┏┓┏┓┃ 11 ┛┗┛┗┛┃ 12 ┓┏┓┏┓┃ 13 ┃┃┃┃┃┃ 14 ┻┻┻┻┻┻ 15 */ 16 #include <algorithm> 17 #include <iostream> 18 #include <iomanip> 19 #include <cstring> 20 #include <climits> 21 #include <complex> 22 #include <fstream> 23 #include <cassert> 24 #include <cstdio> 25 #include <bitset> 26 #include <vector> 27 #include <deque> 28 #include <queue> 29 #include <stack> 30 #include <ctime> 31 #include <set> 32 #include <map> 33 #include <cmath> 34 using namespace std; 35 #define fr first 36 #define sc second 37 #define cl clear 38 #define BUG puts("here!!!") 39 #define W(a) while(a--) 40 #define pb(a) push_back(a) 41 #define Rlf(a) scanf("%lf", &a); 42 #define Rint(a) scanf("%d", &a) 43 #define Rll(a) scanf("%I64d", &a) 44 #define Rs(a) scanf("%s", a) 45 #define Cin(a) cin >> a 46 #define FRead() freopen("in", "r", stdin) 47 #define FWrite() freopen("out", "w", stdout) 48 #define Rep(i, len) for(int i = 0; i < (len); i++) 49 #define For(i, a, len) for(int i = (a); i < (len); i++) 50 #define Cls(a) memset((a), 0, sizeof(a)) 51 #define Clr(a, x) memset((a), (x), sizeof(a)) 52 #define Full(a) memset((a), 0x7f7f, sizeof(a)) 53 #define lrt rt << 1 54 #define rrt rt << 1 | 1 55 #define pi 3.14159265359 56 #define RT return 57 #define lowbit(x) x & (-x) 58 #define onenum(x) __builtin_popcount(x) 59 typedef long long LL; 60 typedef long double LD; 61 typedef unsigned long long ULL; 62 typedef pair<int, int> pii; 63 typedef pair<string, int> psi; 64 typedef map<string, int> msi; 65 typedef vector<int> vi; 66 typedef vector<LL> vl; 67 typedef vector<vl> vvl; 68 typedef vector<bool> vb; 69 70 inline bool scan_d(int &num) { 71 char in;bool IsN=false; 72 in=getchar(); 73 if(in==EOF) return false; 74 while(in!=‘-‘&&(in<‘0‘||in>‘9‘)) in=getchar(); 75 if(in==‘-‘){ IsN=true;num=0;} 76 else num=in-‘0‘; 77 while(in=getchar(),in>=‘0‘&&in<=‘9‘){ 78 num*=10,num+=in-‘0‘; 79 } 80 if(IsN) num=-num; 81 return true; 82 } 83 84 const int mod = 10007; 85 const int maxn = 100010; 86 87 LL add[maxn<<2], put[maxn<<2], mul[maxn<<2]; 88 LL sum1[maxn<<2], sum2[maxn<<2], sum3[maxn<<2]; 89 90 void pushUP(int rt) { 91 sum1[rt] = (sum1[lrt] + sum1[rrt]) % mod; 92 sum2[rt] = (sum2[lrt] + sum2[rrt]) % mod; 93 sum3[rt] = (sum3[lrt] + sum3[rrt]) % mod; 94 } 95 96 void pushDOWN(int rt, int m) { 97 if(put[rt]) { 98 put[lrt] = put[rrt] = put[rt]; 99 add[lrt] = add[rrt] = 0; 100 mul[lrt] = mul[rrt] = 1; 101 sum1[lrt] = (m - (m >> 1)) % mod * put[rt] % mod; 102 sum1[rrt] = (m >> 1) % mod * put[rt] % mod; 103 sum2[lrt] = (m - (m >> 1)) % mod * put[rt] % mod * put[rt] % mod; 104 sum2[rrt] = (m >> 1) % mod * put[rt] % mod * put[rt] % mod; 105 sum3[lrt] = (m - (m >> 1) % mod) * (put[rt] * put[rt]) % mod * put[rt] % mod % mod; 106 sum3[rrt] = (m >> 1) % mod * put[rt] * put[rt] % mod * put[rt] % mod % mod; 107 put[rt] = 0; 108 } 109 if(mul[rt] != 1) { 110 mul[lrt] = mul[lrt] * mul[rt] % mod; 111 mul[rrt] = mul[rrt] * mul[rt] % mod; 112 if(add[lrt]) add[lrt] = (add[lrt] * mul[rt]) % mod; 113 if(add[rrt]) add[rrt] = (add[rrt] * mul[rt]) % mod; 114 sum1[lrt] = (sum1[lrt] * mul[rt]) % mod; 115 sum1[rrt] = (sum1[rrt] * mul[rt]) % mod; 116 sum2[lrt] = (sum2[lrt] * mul[rt]) % mod * mul[rt] % mod; 117 sum2[rrt] = (sum2[rrt] * mul[rt]) % mod * mul[rt] % mod; 118 sum3[lrt] = (sum3[lrt] * mul[rt]) % mod * mul[rt] % mod * mul[rt] % mod; 119 sum3[rrt] = (sum3[rrt] * mul[rt]) % mod * mul[rt] % mod * mul[rt] % mod; 120 mul[rt] = 1; 121 } 122 if(add[rt]) { 123 add[lrt] = (add[lrt] + add[rt]) % mod; 124 add[rrt] = (add[rrt] + add[rt]) % mod; 125 sum3[lrt] = (sum3[lrt] + ((add[rt] * add[rt] % mod) * add[rt] % mod * (m - (m >> 1)) % mod) + 3 * add[rt] * ((sum2[lrt] + sum1[lrt] * add[rt]) % mod)) % mod; 126 sum3[rrt] = (sum3[rrt] + ((add[rt] * add[rt] % mod) * add[rt] % mod * (m >> 1) % mod) + 3 * add[rt] * ((sum2[rrt] + sum1[rrt] * add[rt]) % mod)) % mod; 127 sum2[lrt] = (sum2[lrt] + ((add[rt] * add[rt] % mod) * (m - (m >> 1)) % mod) + (2 * sum1[lrt] * add[rt] % mod)) % mod; 128 sum2[rrt] = (sum2[rrt] + (((add[rt] * add[rt] % mod) * (m >> 1)) % mod) + (2 * sum1[rrt] * add[rt] % mod)) % mod; 129 sum1[lrt] = (sum1[lrt] + (m - (m >> 1)) * add[rt]) % mod; 130 sum1[rrt] = (sum1[rrt] + (m >> 1) * add[rt]) % mod; 131 add[rt] = 0; 132 } 133 } 134 135 void build(int l, int r, int rt) { 136 add[rt] = put[rt] = 0; mul[rt] = 1; 137 sum1[rt] =sum2[rt] = sum3[rt] = 0; 138 if(l == r) return; 139 int m = (l + r) >> 1; 140 build(l, m, lrt); 141 build(m+1, r, rrt); 142 } 143 144 void update(int L, int R, int c, int ch, int l, int r, int rt) { 145 if(L <= l && r <= R) { 146 if(ch == 3) { 147 put[rt] = c; 148 add[rt] = 0; 149 mul[rt] = 1; 150 sum1[rt] = ((r - l + 1) * c) % mod; 151 sum2[rt] = (((r - l + 1) * c) % mod * c) % mod; 152 sum3[rt] = ((((r - l + 1) * c) % mod * c) % mod * c) % mod; 153 } 154 if(ch == 2) { 155 mul[rt] = (mul[rt] * c) % mod; 156 if(add[rt]) add[rt] = (add[rt] * c) % mod; 157 sum1[rt] = (sum1[rt] * c) % mod; 158 sum2[rt] = ((sum2[rt] * c) % mod * c) % mod; 159 sum3[rt] = (((sum3[rt] * c) % mod * c) % mod * c) % mod; 160 } 161 if(ch == 1) { 162 add[rt] += c; 163 sum3[rt] = (sum3[rt] + (((c * c) % mod * c) % mod * (r - l + 1)) % mod + 3 * c * ((sum2[rt] + sum1[rt] * c) % mod)) % mod; 164 sum2[rt] = (sum2[rt] + (c * c % mod * (r - l + 1) % mod) + 2 * sum1[rt] * c) % mod; 165 sum1[rt] = (sum1[rt] + (r - l + 1) * c) % mod; 166 } 167 return; 168 } 169 pushDOWN(rt, r-l+1); 170 int m = (l + r) >> 1; 171 if(R <= m) update(L, R, c, ch, l, m, lrt); 172 else if(L > m) update(L, R, c, ch, m+1, r, rrt); 173 else { 174 update(L, R, c, ch, l, m, lrt); 175 update(L, R, c, ch, m+1, r, rrt); 176 } 177 pushUP(rt); 178 } 179 180 LL query(int L, int R, int p, int l, int r, int rt) { 181 if(L <= l && r <= R) { 182 if(p == 1) return sum1[rt] % mod; 183 if(p == 2) return sum2[rt] % mod; 184 if(p == 3) return sum3[rt] % mod; 185 } 186 pushDOWN(rt, r-l+1); 187 int m = (l + r) >> 1; 188 if(R <= m) return query(L, R, p, l, m, lrt); 189 else if(m < L) return query(L, R, p, m+1, r, rrt); 190 else return (query(L, R, p, l, m, lrt) + query(L, R, p, m+1, r, rrt)) % mod; 191 } 192 193 194 int n, m; 195 int a, b, c, ch; 196 197 int main() { 198 // FRead(); 199 while(~scan_d(n) && ~scan_d(m) && n + m) { 200 build(1, n, 1); 201 W(m) { 202 scan_d(ch); scan_d(a); scan_d(b); scan_d(c); 203 if(ch != 4) update(a, b, c, ch, 1, n, 1); 204 else cout << query(a, b, c, 1, n, 1) << endl; 205 } 206 } 207 RT 0; 208 }
时间: 2024-10-11 05:33:07